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lara31 [8.8K]
2 years ago
11

A ball is dropped from rest out of a high window in a tall building for 5 seconds. Assuming we ignore air resistance and assume

upwards to be positive. A) What will be the final velocity of the ball B) What is the height of the building if it hits the ground after those 5 seconds.
Physics
2 answers:
EleoNora [17]2 years ago
6 0

Answer:

final velocity=-49m/s and height=122.5

Explanation:

u=0,t=5s,v=?

v=u+gt

v=0+(-9.8)5

v=-49m/s

h=ut+1/2gt*2

h=0(5)±1/2(-9.8)5*2

h=-122.5 or 122.5

but height=122.5 because height cannot be negative

lara [203]2 years ago
5 0

Answer:

what is the upwards force?

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Korolek [52]

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3 Draw energy transfer diagrams
timofeeve [1]

Answer:

The outline of the energy transfer are;

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Please find attached the drawings of the energy transfer created with MS Visio

Explanation:

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a) The energy transfer diagram for the winding up a clockwork car is given as follows;

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Kinetic energy → Clockwork spring → Potential energy

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Therefore, we have;

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6 0
2 years ago
8. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bott
nasty-shy [4]

<u>Answer:</u> The ball is travelling with a speed of 5.5 m/s after hitting the <u>bottle.</u>

<u>Explanation:</u>

To calculate the speed of ball after the collision, we use the equation of law of conservation of momentum, which is given by:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m_1,u_1\text{ and }v_1 are the mass, initial velocity and final velocity of ball.

m_2,u_2\text{ and }v_2 are the mass, initial velocity and final velocity of bottle.

We are given:

m_1=0.4kg\\u_1=18m/s\\v_1=?m/s\\m_2=0.2kg\\u_2=0m/s\\v_2=25m/s

Putting values in above equation, we get:

(0.4\times 18)+(0.2\times 0)=(0.4\times v_1)+(0.2\times 25)\\\\v_1=5.5m/s

Hence, the ball is travelling with a speed of 5.5 m/s after hitting the bottle.

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